A capacitor of capacitance 900,mu F is charge | vedclass

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Question

$C =900\,\mu F$

$Q = CV =900 	imes 10^{-6} 	imes 100=9 	imes 10^{-2}=90\,MC$

Now

Common potential will be developed across both capacitors

Vedclass · Accepted Answer

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Vedclass · Answer

$C =900\,\mu F$

$Q = CV =900 	imes 10^{-6} 	imes 100=9 	imes 10^{-2}=90\,MC$

Now

Common potential will be developed across both capacitors by $kVL$

Total charge on left plates of capacitors should be conserved.

$90\,mc +0=2\,cv _0$

$cv _0=45\,mc$

Heat dissipated $= U _{ i }- U _{ f }$ [Change in energy stored in the capacitors]

$=\frac{1}{2} \frac{(90\,mc )^2}{900\,\mu F }-2 	imes \frac{1}{2} \frac{(45\,mc )^2}{900\,\mu F }\left[ U =\frac{ Q ^2}{2 c }ight]$

$=\frac{1}{2 	imes 900 	imes 10^{-6}}(8100-4050) 	imes 10^{-6}$

$=2.25\,Joule$

OR

Heat $=\frac{1}{2} \frac{ C _1 C _2}{ C _1+ C _2}\left( V _1- V _2ight)^2$

$=\frac{1}{2} \frac{ C ^2}{2 C }(100-0)^2$

$=\frac{1}{2} \frac{900 	imes 10^{-6}}{2} 	imes 10^4=\frac{9}{4} 	ext { Joule }=2.25 	ext { Joule }$

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